Question: Simplify and expand the following expression: $ \dfrac{2x}{x + 8}-\dfrac{x - 8}{2x - 5} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(x + 8)(2x - 5)$ Multiply the first term by $\dfrac{2x - 5}{2x - 5}$ $ \begin{align*} \dfrac{2x}{x + 8} \times \dfrac{2x - 5}{2x - 5} & = \dfrac{(2x)(2x - 5)}{(x + 8)(2x - 5)} \\ & = \dfrac{4x^2 - 10x}{(x + 8)(2x - 5)}\end{align*} $ Multiply the second term by $\dfrac{x + 8}{x + 8}$ $ \begin{align*} \dfrac{x - 8}{2x - 5} \times \dfrac{x + 8}{x + 8} & = \dfrac{(x - 8)(x + 8)}{(2x - 5)(x + 8)} \\ & = \dfrac{x^2 - 64}{(2x - 5)(x + 8)}\end{align*} $ Now we have: $ = \dfrac{4x^2 - 10x}{(x + 8)(2x - 5)} - \dfrac{x^2 - 64}{(2x - 5)(x + 8)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{4x^2 - 10x - (x^2 - 64)}{(x + 8)(2x - 5)} $ $ = \dfrac{4x^2 - 10x - x^2 + 64}{(x + 8)(2x - 5)} $ $ = \dfrac{3x^2 - 10x + 64}{(x + 8)(2x - 5)}$ Expand the denominator: $ = \dfrac{3x^2 - 10x + 64}{2x^2 + 11x - 40}$